Closed form solution for $\int_0^{\infty } \frac{\sin ({n}/{x})}{e^{2 \pi x}-1} \, dx$ Is there a closed form solution for the following  integral 
$$\int_0^{\infty } \frac{\sin \left(\frac{n}{x}\right)}{e^{2 \pi  x}-1} \, dx$$
for $n>0$
 A: The proposed integral can be represented as a series, using a Mellin convolution technique. From the inverse Mellin transform of the zeta function:
\begin{equation}
 \frac{1}{e^{x}-1}=\frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty}\Gamma(s)\zeta(s)x^{-s}\,ds
\end{equation} 
where $\Re(c)>1$, one can express
\begin{align}
 I&=\int_0^{\infty } \frac{\sin \left(\frac{n}{t}\right)}{e^{2 \pi  t}-1} \, dt\\
 &=\frac{1}{2\pi}\int_0^{\infty } \frac{\sin \left(\frac{2\pi n}{x}\right)}{e^{x}-1} \, dx\\
 &=\frac{1}{2\pi}\frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty}\Gamma(s)\zeta(s)\,ds\int_0^{\infty }\sin \left(\frac{2\pi n}{x}\right)x^{-s}\,dx\\
 &=\frac{1}{2\pi}\frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty}\left( 2\pi n \right)^{1-s}\Gamma(s)\zeta(s)\,ds\int_0^{\infty }\sin \left(y\right)y^{s-2}\,dy\\
\end{align}
(substitution $y=2\pi n/x$ was made  to obtain the latter expression). Using the Mellin transform for the sine function, valid for $0<\Re(s)<2$, one can express
\begin{equation}
 I=-\frac{1}{2\pi}\frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty}\Gamma(s)\Gamma(s-1)\zeta(s)\cos\left( \frac{\pi}{2}s \right)\left( 2\pi n \right)^{1-s}\,ds
\end{equation} 
where $1<\Re(c)<2$. In the region $\Re(s)<c$, the integrand 
\begin{equation}
 f(s)=\Gamma(s)\Gamma(s-1)\zeta(s)\cos\left( \frac{\pi}{2}s \right)\left( 2\pi n \right)^{1-s}
\end{equation} has poles at $s=1,0$ and $s=-p$, with $p=1,2,3\ldots$ We have
\begin{align}
 \operatorname{Res}\left[ f(s),s=1\right]&=-\frac{\pi}{2}\\
 \operatorname{Res}\left[ f(s),s=0\right]&=-\pi n\left( \ln n+2\gamma-1 \right)
\end{align}
For $s\ne 0,1$, using the reflection formula
\begin{equation}
 \zeta\left(1-s\right)=2(2\pi)^{-s}\cos\left(\tfrac{1}{2}\pi s\right)\Gamma%
\left(s\right)\zeta\left(s\right)
\end{equation} 
the integrand can be written as
\begin{equation}
 f(s)=\pi\Gamma(s-1)\zeta(1-s)n^{1-s}
\end{equation} 
and thus
\begin{equation}
 \operatorname{Res}\left[ f(s),s=-p\right]=(-1)^{p+1}\pi\frac{\zeta(p+1)}{(p+1)!}n^{p+1}
\end{equation} 
Finally, denoting $\gamma$ the Euler's constant,
\begin{equation}
 I=\frac{1}{4}+\frac{n}{2}\left( \ln n+2\gamma-1 \right)+\frac{1}{2}\sum_{p=1}^\infty(-1)^{p}\frac{\zeta(p+1)}{(p+1)!}n^{p+1}
\end{equation} 
Another series expansion can be obtained by developing the denominator of the integrand:
\begin{align}
 I&=\frac{1}{2\pi}\sum_{p=1}^\infty\int_0^\infty e^{-px}\sin\frac{2\pi n}{x}\,dx\\
 &=\frac{1}{2\pi}\sum_{p=1}^\infty\int_0^\infty e^{-\frac{p}{t}}\sin2\pi nt\,\frac{dt}{t^2}
\end{align}
The Fourier sine transform is tabulated (Ederly Vol.1 2.4.30 p.75) (alternatively it can be derived using an integral representation of the modified Bessel function $K_1$):
\begin{equation}
 I=-\sqrt{\frac{2n}{\pi}}\sum_{p=1}^\infty\frac{1}{\sqrt{p}}\Im\left[e^{i\pi/4}K_1 \left(2\sqrt{2\pi np} e^{i\pi/4} \right)\right]
\end{equation} 
It can be noticed that the equality of both series looks very similar to some Ramanujan's summation formulas involving Bessel functions and their generalization (see for example here). 
